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I'm reading Kowalski's Representation theory, and there's a part about the symmetric and antisymmetric powers of a representation, and I'd like to ask a question about those.

So there's a proposition in the book that says if we have a representation $\rho: G \to GL(E)$ and a covariant functor $T:E \mapsto T(E)$, $\Phi \mapsto \Phi_{*}$, from $Vect_{k}$ to itself, then we can construct a representation $T(\rho): G \to GL(T(E))$ naturally. This is obvious and I completely understand this statement.

Then, when defining the symmetric and antisymmetric powers of a representation, the author simply states that there is a covariant functor that maps $E$ to $\mathrm{Sym}^{m}(E)$ (which is, as far as I understand, the set of all symmetric $m$-linear maps $f: E \times \dots \times E \to k$), and thus we can define the $m$-th symmetric power of a representation $\rho: G \to GL(E)$, calling it for example $\mathrm{Sym}^m \rho: G \to GL(\mathrm{Sym}^{m}(E))$. However, I don't know of any covariant functors from $E$ to $\mathrm{Sym}^{m}(E)$; all I know is a contravariant functor: $$ E \mapsto \mathrm{Sym}^{m}(E), (\Phi: E_{1} \to E_{2}) \mapsto (\Phi^{*}: \mathrm{Sym}^{m}(E_{2}) \to \mathrm{Sym}^{m}(E_{1}))$$

given by $\Phi^{*}(f)(v_{1},...,v_{m}) = f(\Phi(v_{1}),...,\Phi(v_{m}))$.

My question is: What is the symmetric power of a representation? Here are some of my observations:

  1. If there were a natural covariant functor from $E$ to $\mathrm{Sym}^{m}(E)$, i.e. if someone knows of one, I'd appreciate a clear formulation of one, which would answer my question.
  2. Is there a way to define a representation with regard to a contravariant tensor? Here's my attempt in this particular case:

$$\mathrm{Sym}^{m}\rho: G \to \mathrm{Sym}^{m}(E), \mathrm{Sym}^{m}\rho(g)f(v_{1},...,v_{m}) = f(\rho(g^{-1})v_{1},...,\rho(g^{-1})v_{m}).$$ This is indeed a homomorphism, and the reason I had to put $g^{-1}$ instead of $g$ was because if I had put $g$, I'd have $\mathrm{Sym}^{m}\rho(gh) = \mathrm{Sym}^{m}\rho(h) \circ \mathrm{Sym}^{m}\rho(g)$. Is this the standard definition?

I would appreciate any details with regards to what is standard in this case.

Matija Sreckovic
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    Your definition of $\operatorname{Sym}^m E$ is off by a "dual", because it is actually the space where all symmetric $m$-linear maps $E\times E\times\dots\times E\to X$ factors through: $E\times E\times\dots\times E\to\operatorname{Sym}^m E\to X$. – user10354138 Nov 20 '18 at 14:41
  • I forgot to put "symmetric" in my definition of $\mathrm{Sym}^{m}(E)$. So the set of all symmetric $m$-linear maps from $E \times E \times \dots \times E$ to $k$ is actually $(\mathrm{Sym}^{m}(E))^{*}$? Could you explain to me the exact definition of $\mathrm{Sym}^{m}(E)$, and how one works with that space (for example, could you name a basis)? Or even better, point me to a book which contains some clear information on the symmetric algebra? – Matija Sreckovic Nov 20 '18 at 15:11
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    If $x_1, \ldots, x_n$ is a basis of $E$, then a standard basis of $\operatorname{Sym}^m(E)$ would be ${ x_{i_1} x_{i_2} \cdots x_{i_m} \mid 1 \le i_1 \le i_2 \le \cdots \le i_m \le n }$ (where $x_{i_1} x_{i_2} \cdots x_{i_m}$ denotes the image of $x_{i_1} \otimes \cdots \otimes x_{i_m}$ in the quotient described in Qiaochu Yuan's answer.) Therefore $\operatorname{Sym}^m(E)$ has dimension $\binom{n+m-1}{m}$. And the symmetric algebra $S(E)$ is isomorphic to the polynomial algebra $F[x_1, \ldots, x_n]$ where $F$ is the scalar field. – Daniel Schepler Nov 21 '18 at 01:13

1 Answers1

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What you've defined is not the symmetric power, since as you say it's contravariant. The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{\otimes n}$ (which is also covariant) by the action of the symmetric group $S_n$ permuting the factors. It has the following universal property: every symmetric multilinear map $V^n \to W$ factors uniquely through a linear map $S^n(V) \to W$. So there is a universal symmetric multilinear map

$$V^n \to S^n(V).$$

What you call the symmetric power is the dual of $S^n(V)$.

The symmetric powers are the graded components of the symmetric algebra

$$S(V) = \bigoplus_{n \ge 0} S^n(V)$$

and in particular there's a bilinear map $S^n(V) \times S^m(V) \to S^{n+m}(V)$ giving the symmetric algebra the structure of a commutative algebra. If $e_1, \dots e_n$ is a basis for $V$, $S^n(V)$ has a basis consisting of all monomials in the $e_i$, thought of as variables, of degree $n$, and $S(V)$ is the polynomial algebra on the $e_i$.

As for your second question, yes, and that's how you do it. See dual representation.

Qiaochu Yuan
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  • Sorry for the silly question, but: what do you exactly mean by "The $n^{th}$ symmetric power $S^n(V)$ of a vector space $V$ is the quotient of the $n^{th}$ tensor power $V^{\otimes n}$ by the action of the symmetric group $S_n$ permuting the factors"? I imagine you are not considering the set-theoretic quotient by the equivalence relation $v \sim w \iff w = g\cdot v$ for some $g \in S_n$, isn't it? – Ender Wiggins Nov 19 '20 at 11:54
  • @Ender: I mean the quotient as a vector space. – Qiaochu Yuan Nov 19 '20 at 17:56